5 Body Collinear Central Configurations. Candice Davis, Scott Geyer, William Johnson
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1 5 Body Collinear Central Configurations Candice Davis, Scott Geyer, William Johnson
2 5 Body Collinear Central Configuration 1. Purpose 2. General Case Equations 3. Symmetric Case 4. r 0 Case (r > 0 and r < 0) 5. r = 0 Case 6. Super Central Configurations
3 The Purpose for this research! Central configurations are one of the most significant and fundamental topics in the study of few-body problems, which is why it gained so much popularity. We generally understand as the number of bodies increase, the problems become more complex. The 2 body problem has been completely solved. The 3-body and 4-body problem have already been solved when given a mass or position, which is why we have chosen to research the 5-body collinear configuration. For a general case when given mass, the 5-body problem has also already been solved, but when given the position, it has not. Hence, our reasoning for this research. We took a deeper look into the 5-body problem on a linear configuration. We have been working with general case which we split into special cases.
4 Three Body Problem The gravitational three-body problem is one of the oldest problems in mathematical physics. The goal is to determine the trajectories of three interacting particles. Historically, the system involving the Moon, the Earth and the Sun was the first three-body problem that received extended study. Since the time of Newton, physicist and mathematicians have applied significant efforts to solve this problem. To this day, it has no known general solution. In 1767, Euler discovered a collinear central configuration. Painting of Euler by Jakob Emanuel
5 Three Body Problem Chaos! There is no analytical solution to the 3-body problem. Can be solved and has been solved numerically. Source: Ask a Mathematician Photo of Poincare by Eugène Pirou
6 Four Body Problem Xie and Ouyang (2005) established an expression for the 4 masses depending on the position of x and the center of mass u, which give central configurations in the 4-body problem. By using the inverse problem, they found possible positive masses while maintaining a central configuration. Image: Hampton, M. UMN Duluth
7 Xie and Ouyang Process for 4-Body - They found that if they fix the center of mass at the origin, there are four unique positive masses that will make the configuration central, if and only if the center of mass is fixed. -Using four curves they found a region in the first quadrant that they refer to as the central configuration region. They can use any point within that region and have 4 positive masses that will make it a central configuration. -If they changed the masses with respect to the position, they found that whether or not they fixed the center of mass they were unable to get four positive masses that formed a central configuration. -They found that the central configuration region is symmetric when s=t, when s=t, m1=m4 and m2=m3, with the center of mass fixed at the origin. This gave results of m1 and m4 going to infinity as s=t goes to infinity and m2 and m3 gave a finite number of 68. Using this it was found in the symmetric region where collinear 4 bodies lies. - We used some of these same approaches in our the symmetric case for the 5-body problem by making our 3rd mass fixed at the center of mass on the origin.
8 Five Body Problem Image: Jupiter and the four moon first discovered by Galileo (Ask a Mathematician)
9 In this research, we studied the central collinear configurations of the 5-body problem. A central configuration for n-body problems is formed if the position vector of each particle with respect to the center of mass is a common scalar multiple of its acceleration. Our goal was to see what information we could derive about the position parameters r, s, and t with masses m1, m2, m3, m4, and m5. We want positive masses that make the configuration central. We wanted to know what the central mass relies on and if there are any positive masses that can form a central configuration. Five masses and their distances were used to help formulate the equations that helped find the center of mass. s>0-1 < r < 1 t>0
10 The formula we used to help establish our equations for our masses is the Newtonian n-body problem: [1] If the acceleration vectors of the bodies satisfy the above equation then the configuration is called a central configuration.
11 General Equations
12 In order to prevent division by zero and elimination of variables, we must first move m3 in the m1, m2, m4, and m5 equations to the right side. To find expressions for m1, m2, m4, m5, we use Gaussian elimination in an augmented matrix.
13 In order to ensure we did not divide by zero in any step, we checked to ensure that the denominator d(s) 0. Since s > 0 and t > 0, (s + 2)2(s + t)2 > s2t2. Therefore, d(s) > 0.
14 Equations for masses 1 and 2
15 Equations for masses 4 and 5
16 Solving for C In order to satisfy the equation for m3, we substitute the expressions for m1, m2, m4, and m5 into the equation for m3 to find an expression of c.
17 Equation for C While the equation for c is rather large, it still depends only on the parameters r, s, and t.
18 Introduce Total Mass and change form of Equations m1,m2,m4 and m5. To generate equations in terms of the total mass, M, we substitute our expressions for m1,m2,m4,m5 into the equation for total mass and solve for m3. We then substitute this expression for m3 into the equations for m1,m2,m4,m5 to get them in the form fiλ + gim.
19 Equation for Total Mass
20 Equations in fiλ+gim form To generate these equations f
21 GeoGebra Test Model
22 Symmetric Case (r = 0, s = t)
23 Symmetric Case Linear graph used to form our equations for masses m1, m2, m3, m4, and m5 s>0
24 Symmetric Region Theorem
25 Furthermore, We Found for the Symmetric Case...
26 Symmetric Case In the symmetric case, we found that m1=m5, m2=m4, and c = 0. Since m1 = m5 and m2 = m4, we can simplify the problem down to two equation as the same parameters that satisfy m1 will satisfy m5 and likewise for m2 and m4. We chose to use m4 and m5 instead of m1 and m2 because of their simplicity. Substituting 0 for r and s for t, we obtain the following equations that appear on the next slide.
27 Symmetric Case Equations In order to have a central configuration, we need positive masses. Therefore, we set m4 > 0 And m5 > 0
28 Symmetric Case Inequalities Solving the inequalities m5 > 0 and m4 > 0 for gives us: h2 denominator (h2d) is zero at s=s ~ Division by h2d splits h2 into two cases.
29 Symmetric Case Region for λ/m3 in terms of s.
30 Symmetric Case GeoGebra Test Model
31 r 0 Case
32 r 0 Region Theorem
33 Formulate Inequalities
34 Rename for Simplicity min1, min3, min5 max2, max4
35 r > 0 Case
36 Fix r As r becomes larger, values of s become unusable. Center of mass moves too far away from center of configuration. Can find a central configuration for a given t regardless of value for r. Bulge near t = 1 which becomes a dent as r becomes larger.
37 Fix t Central configurations are possible below the line and not possible above the line.
38 Fix t Central configurations are possible below the line and not possible above the line.
39 t = 1 Region Outlines
40 t = 1 Regions Conditions for central configuration.
41 t = 1 Regions Always possible in regions max4 > min3, max2 > min1, max4 > min5, and max2 > min5 The region is bounded by the implicit curve min1 = max4
42 Implicit plot of min1 = max4 for given values of t
43 Implicit plot of min1 = max4 for given values of t
44 Implicit plot of min1 = max4 for given values of t Boundary for possible central configurations for given values of t
45 simplified expression for min1 < max4
46 In Terms of Center of Mass In order to be a central configuration, the center of mass cannot move too far away from the center of the configuration.
47 r > 0 Case GeoGebra Test Model
48 r < 0 case
49 r < 0 Case The r < 0 case is a rotation of the r > 0 case and is considered the same configuration.
50 r < 0 Case We can swap s and t, m1 and m5, m2 and m4 and achieve the same results. min5 < max2 in the s = 1 plane
51 r < 0 Case Implicit plot of min5 = max2 superimposed on region where central configurations are possible on the s = 1 plane.
52 r > 0 Case r < 0 Case min1 = max4 min5 = max2 r r s t s t
53 r = 0 Case
54 r = 0 Region Theorem
55 r=0 Outline of Regions.
56 r = 0 Regions By rotation, we can reduce the number of regions from 6 to 3. Central configuration is possible in all regions. When r = 0, no areas were found where central configurations were not possible.
57 Conclusion For r > 0: central configurations are possible under the curve min1 = max4. For r < 0: central configurations are possible above the curve min5 = max2. For r = 0: central configurations are possible for all s and t. Future Plans Further analyze the regions and conditions to develop better understanding. Work on Super Central Configurations, which analyzes the permutations of the masses of central configurations.
58 Acknowledgement A special thanks to Dr. Xie, The University of Southern Mississippi Mathematics Department and the Wright W. and Annie Rea Cross Foundation scholarship for making this research possible.
59 References L. Euler, De motu rectilineo trium corporum se mutuo attahentium, Movi Comm. Acad. Sci. Imp. Petrop. 11 (1767), Ouyang, T., Xie, Z., Collinear central configuration in four-body problem. Celest. Mech. Dyn Astron. 93, (2005). The Physicist, What is the three-body problem?, Ask a mathematician.
60 Images J. Emanuel, Painting of Euler. Retrieved from wikimedia.org E. Pirou, Photo of Poincare. Retrieved from wikimedia.org M. Hampton, four body collinear orbit, University of Minnesota Duluth The Physicist. Jupiter and four of its moons. Retrieved from Ask a Mathematician The Physicist. Three body equation. Retrieved from Ask a Mathematician
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